Prove that the sum of the 'pressure head', 'velocity head' and the gravitational
head' remains constant during the stream line flow of an ideal liquid
Answers
Its an Bernoulli Theorem.
According to the bernoulli theorem, the sum of pressure head, gravitational head and velocity head always remains the constant.
Proof of this theorem is shown in the attachment.
Assumption required for this theroem are ---
1. Fluid should be incompressible.
2. Fluid should have streamline flow.
3. Fluid should be irrotational.
Now refers to the proof which is shown in attachment.
Hope the proof will help you.
ANSWER:-
Consider a fluid of negligible viscosity moving with laminar flow
Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2. If the fluid is incompressible:
A1L1 = A2L2
The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:
Work done = force x distance = p x volume
Net work done per unit volume = P1 - P2
k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume)
Therefore:
k.e. gained per unit volume = ½ ρ(v22 - v12)
p.e. gained per unit volume = ρg(h2 – h1)
where h1 and h2 are the heights of Q and R above some reference level. Therefore:
P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1)
P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2
Therefore:
P + ½ ρv2 + ρgh is a constant
For a horizontal tube h1 = h2 and so we have:
P + ½ ρv2 = a constant
This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.
No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.
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